R50 · Thin lens equation: image distance and magnification

This problem uses the thin lens equation to determine the image characteristics.

### Given Values:
*   Focal length ($f$) = $+10 \text{ cm}$ (Positive because it is a converging lens)
*   Object distance ($d_o$) = $15 \text{ cm}$

---

## (a) Calculate the Image Distance ($d_i$)

We use the thin lens equation:
$$\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$$

Rearrange the equation to solve for $\frac{1}{d_i}$:
$$\frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o}$$

Substitute the known values:
$$\frac{1}{d_i} = \frac{1}{10 \text{ cm}} - \frac{1}{15 \text{ cm}}$$

To subtract the fractions, find a common denominator (which is 30):
$$\frac{1}{d_i} = \left(\frac{3}{30 \text{ cm}}\right) - \left(\frac{2}{30 \text{ cm}}\right)$$
$$\frac{1}{d_i} = \frac{3 - 2}{30 \text{ cm}}$$
$$\frac{1}{d_i} = \frac{1}{30 \text{ cm}}$$

Invert the result to find $d_i$:
$$d_i = +30 \text{ cm}$$

**The image distance ($d_i$) is $+30 \text{ cm}$.**

---

## (b) Calculate the Magnification ($m$)

We use the magnification equation:
$$m = -\frac{d_i}{d_o}$$

Substitute the values for $d_i$ and $d_o$:
$$m = -\frac{(+30 \text{ cm})}{(15 \text{ cm})}$$
$$m = -2$$

**The magnification ($m$) is $-2$.**

---

## Conclusion and Image Characteristics

| Characteristic | Value | Interpretation |
| :--- | :--- | :--- |
| **Image Distance ($d_i$)** | $+30 \text{ cm}$ | Positive sign means the image forms on the opposite side of the lens from the object. The image is **Real**. |
| **Magnification ($m$)** | $-2$ | Negative sign indicates that the image is flipped relative to the object. The image is **Inverted**. |

The resulting image is **REAL** and **INVERTED**. (It is also magnified, as $|m|=2$).

To solve the problem, we will apply the

To solve this problem, we will use the given values and the thin lens equation.

### **Given Data:**
*   Focal length ($f$) = $+10\text{ cm}$ (positive because it is a converging lens)
*   Object distance ($d_o$) = $15\text{ cm}$

---

### **(a) Calculate the image distance ($d_i$)**

Using the thin lens equation:
$$\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$$

Rearrange the equation to solve for $\frac{1}{d_i}$:
$$\frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o}$$

Substitute the given values:
$$\frac{1}{d_i} = \frac{1}{10} - \frac{1}{15}$$

To subtract these fractions, find a common denominator (which is 30):
$$\frac{1}{d_i} = \frac{3}{30} - \frac{2}{30}$$
$$\frac{1}{d_i} = \frac{1}{30}$$

Invert both sides to find $d_i$:
**$d_i = 30\text{ cm}$**

---

### **(b) Calculate the magnification ($m$)**

Using the magnification formula:
$$m = -\frac{d_i}{d_o}$$

Substitute the values:
$$m = -\frac{30\text{ cm}}{15\text{ cm}}$$
**$m = -2$**

---

### **Final Image Characteristics**

Based on the calculations above, we can determine the nature of the image:

1.  **Real or Virtual?** Since the image distance ($d_i$) is **positive** ($+30\text{ cm}$), the image is **real**. (Real images are formed on the opposite side of the lens from the object).
2.  **Upright or Inverted?** Since the magnification ($m$) is **negative** ($-2$), the image is **inverted**.
3.  **Size:** Since the absolute value of the magnification is greater than 1 ($|m| > 1$), the image is **enlarged** (twice the size of the object).

**Summary:** The image is **real, inverted, and enlarged**.

This problem uses the thin lens equation to determine the location and characteristics of an image formed by a convex lens.

### Given Values:
*   Focal length ($f$): $+10 \text{ cm}$ (Positive for a converging lens)
*   Object distance ($d_o$): $+15 \text{ cm}$
*   Thin Lens Equation: $\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$

***

### (a) Calculate the image distance $d_i$

We rearrange the thin lens equation to solve for $\frac{1}{d_i}$:
$$\frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o}$$

Substitute the given values:
$$\frac{1}{d_i} = \frac{1}{10 \text{ cm}} - \frac{1}{15 \text{ cm}}$$

To subtract the fractions, find a common denominator (which is 30):
$$\frac{1}{d_i} = \left(\frac{3}{30}\right) - \left(\frac{2}{30}\right)$$
$$\frac{1}{d_i} = \frac{3 - 2}{30}$$
$$\frac{1}{d_i} = \frac{1}{30 \text{ cm}}$$

Invert the expression to find $d_i$:
$$d_i = +30 \text{ cm}$$

**The image distance ($d_i$) is $+30 \text{ cm}$.**

***

### (b) Calculate the magnification $m$

We use the magnification formula:
$$m = -\frac{d_i}{d_o}$$

Substitute the calculated and given values:
$$m = -\frac{(+30 \text{ cm})}{(+15 \text{ cm})}$$
$$m = -2$$

**The magnification ($m$) is $-2$.**

***

### Conclusion and Image Characteristics

| Parameter | Value | Interpretation |
| :--- | :--- | :--- |
| **Image Distance ($d_i$)** | $+30 \text{ cm}$ | Since $d_i$ is positive, the image forms on the opposite side of the lens from the object. $\implies$ **Real** |
| **Magnification ($m$)** | $-2$ | The negative sign indicates the image is inverted. $|m|=2$ means it is twice as large. $\implies$ **Inverted** |

The resulting image is a **real**, **inverted**, and magnified image, located $30 \text{ cm}$ from the lens on the side opposite the object.

## Step 1: Given information
The focal length of the lens (f) is given as 10 cm. The object distance (d_o) is given as 15 cm.

## Step 2: Using the thin lens equation to find d_i
We use the formula 1/f = 1/d_o + 1/d_i, where f = 10 cm and d_o = 15 cm.

## Step 3: Substitute values into the thin lens equation
Substituting the given values into the thin lens equation gives us 1/10 = 1/15 + 1/d_i.

## Step 4: Solve for d_i
To solve for d_i, we first need to get a common denominator on the right side of the equation. This results in (d_o * d_i) / (d_o * d_i) being equal to 1/10 = 1/15 + 1/d_i.

## Step 5: Multiply both sides by d_o * d_i
After multiplying, we get d_i / (d_o * d_i) = 3*d_i / (15*d_i) + (d_o*d_i) / (d_o*d_i), which simplifies to 1/10 = 1/15 + 1/d_i.

## Step 6: Find a common denominator and simplify
Finding a common denominator of 150 for both fractions on the right side, we get 15/d_i = 10/d_i + 10/150.

## Step 7: Combine like terms
Combining the terms gives us 15/d_i = 10/d_i + 1/15*d_i. Multiplying everything by d_i to clear the fraction, we get 15 = 10 + (d_i)/15.

## Step 8: Solve for d_i
Subtracting 10 from both sides, we have 5 = (d_i)/15.

## Step 9: Multiply both sides by 15
Multiplying both sides of the equation by 15 gives us d_i = 5*15.

## Step 10: Calculate d_i value
The calculation results in a positive value for d_i, indicating that the image is real. The calculated d_i is 75 cm.

## Step 11: Determine magnification and orientation
Since d_i is positive, the image is real. The magnification m = -d_i/d_o, so we calculate it as follows: m = -(75) / (15).

## Step 12: Calculate magnification value
The calculation results in a negative value for magnification, indicating that the image is inverted.

## Step 13: State whether image is real or virtual and upright or inverted
Based on the calculations, we conclude that the resulting image is real and inverted.


The final answer is: $\boxed{75}$

This problem uses the thin lens equation to determine where an image forms and what its characteristics are.

### Given Values:
*   Focal length ($f$): $+10 \text{ cm}$ (Positive because it is a converging lens)
*   Object distance ($d_o$): $15 \text{ cm}$
*   Thin Lens Equation: $\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$

---

### (a) Calculate the image distance $d_i$

We rearrange the thin lens equation to solve for $1/d_i$:
$$\frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o}$$

Substitute the given values:
$$\frac{1}{d_i} = \frac{1}{10 \text{ cm}} - \frac{1}{15 \text{ cm}}$$

To subtract these fractions, we find a common denominator (which is 30):
$$\frac{1}{d_i} = \left(\frac{3}{30}\right) - \left(\frac{2}{30}\right)$$
$$\frac{1}{d_i} = \frac{3 - 2}{30}$$
$$\frac{1}{d_i} = \frac{1}{30}$$

$$d_i = +30 \text{ cm}$$

**The image distance $d_i$ is $+30 \text{ cm}$.**

---

### (b) Calculate the magnification $m$

We use the formula:
$$m = -\frac{d_i}{d_o}$$

Substitute the calculated $d_i$ and the given $d_o$:
$$m = - \frac{30 \text{ cm}}{15 \text{ cm}}$$
$$m = -2$$

**The magnification $m$ is $-2$.**

---

### Image Classification

Based on the signs of $d_i$ and $m$:

1.  **Real or Virtual?**
    *   Since $d_i$ is positive ($+30 \text{ cm}$), the image forms on the same side as the object (the opposite side of the lens). Therefore, the image is **real**.

2.  **Upright or Inverted?**
    *   The magnification $m$ is negative ($-2$). A negative magnification indicates that the image is flipped relative to the object. Therefore, the image is **inverted**.

*(Note: The magnitude $|m|=2$ also tells us that the image is twice the size of the original object.)*